Two steel bars carry the same tensile force of 2000 N. Bar A has a cross-sectional area of 10 mm² and Bar B has an area of 200 mm². Which statement is correct?
ABoth bars experience the same stress because the force is identical
BBar A has 20 times the stress of Bar B, and is far more likely to yield first
CBar B has greater stress because it has more material resisting the load
DStress cannot be compared between bars of different sizes
Stress = F/A. Bar A: σ = 2000/10 = 200 MPa; Bar B: σ = 2000/200 = 10 MPa. Bar A has 20× the stress. This is the central insight: stress normalizes force by area, which is why it predicts material failure while raw force does not. Option A is the classic misconception — equal forces do not mean equal stress. Option C reverses the logic: more area means the force is distributed over more material, reducing the intensity.
Question 2 Multiple Choice
A rod originally 500 mm long stretches to 503 mm under load. What is the engineering strain?
A3 mm
B0.006 (dimensionless)
C3/503 ≈ 0.00596
D500/503 ≈ 0.994
Engineering strain ε = ΔL / L₀ = (503 − 500) / 500 = 3/500 = 0.006. It is dimensionless — a fraction, not a length. Option A gives the raw elongation in mm, which is not strain. Option C uses the deformed length L in the denominator — that would be true strain (approximately equal for small deformations, but not the engineering definition). Option D gives the ratio of original to deformed length, which has no standard physical meaning here.
Question 3 True / False
True stress and engineering stress are equal regardless of how much a material has deformed.
TTrue
FFalse
Answer: False
Engineering stress uses the original cross-sectional area A₀, which stays fixed in the formula even as the material deforms. True stress uses the instantaneous area A, which decreases as the material stretches (due to Poisson's ratio: lateral contraction accompanies axial extension). For small deformations in the elastic range, the two are approximately equal. Once plastic deformation begins — especially past the necking point — the cross-section shrinks significantly, and true stress diverges above engineering stress. The two can differ by 30–50% at large strains.
Question 4 True / False
Stress is a more useful measure than applied force for predicting whether a material will yield because stress accounts for how the force is distributed across the cross-section.
TTrue
FFalse
Answer: True
This is the fundamental reason stress was defined in the first place. A thin wire carrying 500 N will snap; a thick cable carrying 500 N will barely notice it. The material 'experiences' not the total force but the force per unit area — the intensity of loading. Yield and fracture criteria (like the von Mises or Tresca criteria) are expressed in terms of stress, not force, precisely because they are material properties that depend on internal loading intensity, not on the structural geometry of the component.
Question 5 Short Answer
Why is stress (force per unit area) a more useful measure for predicting material failure than the total applied force?
Think about your answer, then reveal below.
Model answer: Total force tells you how hard you are pulling on an object, but not how intensely that pull is distributed through the material. A small cross-section under modest force can be on the verge of failure while a large cross-section under a much greater force is well within its safe range. Stress normalizes the force by the area, yielding a measure of internal loading intensity that depends on the material's own properties — yield strength, ultimate strength — rather than on the geometry of the specific component. Failure criteria are threshold stresses, not threshold forces.
This is why material datasheets report yield strength and ultimate tensile strength in MPa (stress units), not Newtons. The same material in any geometry fails at the same critical stress level. Force-based predictions would require a separate characterization for every possible cross-section shape and size.